Suppose instead of looking at the rotation of an entire merry-go-round, we examine only the change in the position of one rider on its surface, shown by the orange dot. For this initial discussion, we are going to assume that the merry-go-round is rotating at a constant rate so that the rider moves through a circular path, or linear circumference, at a constant speed. Please be conscious of the fact that the rider's velocity is not constant since the direction of her motion is constantly changing as shown in the second diagram.

As long as the rider does not move along the merry-go-round's surface, the magnitude of the rider's velocity (that is, her speed) equals where is the merry-go-round's constant angular velocity and r is the rider's radial distance from the center of the merry-go-round.

Although the merry-go-round has no angular acceleration, the rider is experiencing a centripetal acceleration towards the center of the circle, or the axis of rotation. This type of acceleration acts parallel to the radius of the circle and is often referred to as radial acceleration.

As with linear acceleration, centripetal acceleration also points in the direction of the change in velocity. Vectorially, this is confirmed in the diagram to the left where you see the vector pointing to the center of the circle. Remember that and that subtracting a vector just means that you add the same magnitude vector pointing in 180º the opposite direction.

This type of acceleration is called uniform centripetal acceleration since the object's speed is not changing, just its direction is changing at a uniform rate based on the merry-go-round's angular velocity.

If the force causing this centripetal (or radial - "along the radius") acceleration were to be removed, the object would "fly off at a tangent" and no longer move in a circular path.

As shown in the previous lesson, an object's centripetal acceleration is calculated with the formula . Since we can rewrite this equation as .

If asked to calculate the rider's total linear acceleration, you would also need to examine any tangential acceleration present, . In our case, the angular acceleration equals zero since we are demanding that the merry-go-round rotate at a constant angular velocity. Since and the rider's net acceleration just equals the center-seeking acceleration along the radius, or .

Now let's examine a rotational system which is experiencing angular acceleration. We know from our previous study of rotational dynamics, that an angular acceleration requires a torque. Our model will be a fixed pulley that has a string wrapped around its perimeter from which a mass is suspended. When the mass is released, a torque will be provided by the tension in the string resulting in a counterclockwise rotation of the pulley.

A freebody diagram of the falling mass would show two forces: T and mg. Since the mass is falling, mg must be greater than the tension, T.

Solving for tension we have

Since the string does not slip over the pulley, but causes it to rotate, we know that .

In this example, a "piece of dust" remaining in place on the rim of the pulley would experience two types of linear acceleration: centripetal acceleration directed towards the center of the pulley and tangential acceleration parallel to the edge of the pulley.

Since these two accelerations are at right angles to each other, the net linear acceleration is calculated using the Pythagorean Theorem.

Also note that this value would be an instantaneous value since the centripetal acceleration is dependent on the instantaneous angular/linear velocity.